solve (dy/dx)+y=1

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

3. solve (dy/dx)+y=1

(dy/dx)+y=1

(dy/dx)=1-y

 

separating the variables

dy / [1-y] =dx

integrating,

[remember to divide by (-1), the coefficient of y]

{ln[1-y] }/(-1)=x+ln C

 

{ here ln refers to natural logarithms }

-ln(1-y) =x+ln C

lnC +ln(1-y) = -x

 

using properties of logarithms

ln [C(1-y)]= (-x)

getting rid of logarathmic funtion

C(1-y)= e^(-x)

(1-y) = (1/C) e^(-x)

y=1-(1/C) e^(-x)

 

put (1/C) =(-A)

 

y = 1 + A e^(-x)

 

 4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 

divide each term by tan y tan x 

[ { (sec x)^2}/ tanx ]dx + [ { (sec y)^2}/ tan y ]dy = 0

integrating term by term 

keeping in mind that integral of {(f ') / f }form is ln(f)

 

{ here ln refers to natural logarithms }

 

ln(tan x) +ln(tan y) =lnC

using properties of logarithms

 

ln(tanx tany ) =lnC

tan x tan y = C

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 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations

solve dy/dx = [1-cosx]/[1+cosx]

solution

2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

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